A spherical balloon is being inflated. Given that the volume of a sphere in terms of its radius is V(r) =4/3 πr^3 and the surface area of a sphere in terms of its radius is S(r) = 4 πr^2, estimate the rate at which the volume of the balloons is changing with respect to its surface area when the surface area measures 50 cm^2.
dV/dS = (dV/dt) / (dS/dt) = (4πr^2 dr/dt)/(8πr dr/dt) = r/2
So just find r when V = 50
Can you show that too
To estimate the rate at which the volume of the balloon is changing with respect to its surface area, we need to find the derivative of the volume function V(r) with respect to the surface area function S(r). This can be done using the chain rule.
First, let's differentiate the volume function V(r) with respect to the radius r:
dV/dr = d/dr(4/3 πr^3)
= 4/3 * 3πr^2
= 4πr^2
Next, let's differentiate the surface area function S(r) with respect to the radius r:
dS/dr = d/dr(4πr^2)
= 8πr
Now, we can find the rate at which the volume is changing with respect to the surface area by dividing dV/dr by dS/dr:
(dV/dr)/(dS/dr) = (4πr^2)/(8πr)
= 1/2r
So, the rate at which the volume of the balloon is changing with respect to its surface area is 1/2r. To find the rate when the surface area measures 50 cm^2, we need to find the corresponding radius of the balloon.
Given that the surface area is S(r) = 4πr^2 = 50 cm^2, we can solve for r:
4πr^2 = 50
r^2 = 50/(4π)
r = √(50/(4π))
Now that we have the radius, we can substitute it into the rate formula:
Rate = 1/2r
= 1/(2 * √(50/(4π)))
Simplifying further gives the estimate for the rate at which the volume is changing with respect to the surface area when the surface area measures 50 cm^2.